A bird is flying north with a mass of 2.5kg has a momentum of 17.5kg m/s at what velocity is it flying
2 answers:
Answer:
v=7 m/s
Explanation:
momentum definition
The amount of movement or linear momentum is the product of velocity by mass. Velocity is a vector while mass is a scalar. As a result we obtain a vector with the same direction and direction as the velocity.
The magnitude of the momentum is calculated like this:
p=m*v Formula (1)
where:
p: momentum (kg*m/s)
m: mass (kg)
v: velocity (m/s)
Data
m = 2.5kg
p = 17.5kg m/s
Problem development
We replace the data in the formula (1)
17.5kg m/s = 2.5kg*v
v = 7 m/s
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Answer:
The work done by friction was
Explanation:
Given that,
Mass of car = 1000 kg
Initial speed of car =108 km/h =30 m/s
When the car is stop by brakes.
Then, final speed of car will be zero.
We need to calculate the work done by friction
Using formula of work done
Put the value of m and v
Hence, The work done by friction was
Answer:
21.8°
Explanation:
Let's call θ the angle between BC and the horizontal.
Draw a free body diagram for each block.
There are 4 forces acting on block D:
Weight force P pulling down,
Normal force N₁ pushing perpendicular to AB,
Friction force N₁μ pushing parallel up AB,
and tension force T pushing parallel up AB.
There are 4 forces acting on block E:
Weight force P pulling down,
Normal force N₂ pushing perpendicular to BC,
Friction force N₂μ pushing parallel to BC,
and tension force T pulling parallel to BC.
Sum of forces on D in the perpendicular direction:
∑F = ma
N₁ − P sin θ = 0
N₁ = P sin θ
Sum of forces on D in the parallel direction:
∑F = ma
T + N₁μ − P cos θ = 0
T = P cos θ − N₁μ
T = P cos θ − P sin θ μ
T = P (cos θ − sin θ μ)
Sum of forces on E in the perpendicular direction:
∑F = ma
N₂ − P cos θ = 0
N₂ = P cos θ
Sum of forces on E in the parallel direction:
∑F = ma
N₂μ + P sin θ − T = 0
T = N₂μ + P sin θ
T = P cos θ μ + P sin θ
T = P (cos θ μ + sin θ)
Set equal:
P (cos θ − sin θ μ) = P (cos θ μ + sin θ)
cos θ − sin θ μ = cos θ μ + sin θ
1 − tan θ μ = μ + tan θ
1 − μ = tan θ μ + tan θ
1 − μ = tan θ (μ + 1)
tan θ = (1 − μ) / (1 + μ)
Plug in values:
tan θ = (1 − 0.4) / (1 + 0.4)
θ = 23.2°
∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.
